A body is projected at an angle of 30° to the horizontal with a velocity of 150 ms\(^{-1.}\). Calculate the time it takes to reach the greatest height [Take...
Question 1 Report
A body is projected at an angle of 30° to the horizontal with a velocity of 150 ms\(^{-1.}\). Calculate the time it takes to reach the greatest height [Take g = 10 ms\(^2\) and neglect air resistance]
Answer Details
At the greatest height the vertical component of velocity is zero. Time to reach the greatest height:
\[ t = \frac{u\sin\theta}{g}. \]With \(u = 150\ \text{m s}^{-1}\), \(\theta = 30^\circ\), \(g = 10\ \text{m s}^{-2}\):
\[ t = \frac{150 \times \sin 30^\circ}{10} = \frac{150 \times 0.5}{10} = \frac{75}{10} = 7.5\ \text{s}. \]Time to reach greatest height \(= 7.5\ \text{s}\).
Download The App On Google Playstore
Everything you need to excel in your exams
Personalized AI Learning Chat Assistant
200,000+ Exam Questions Across IGCSE, JAMB, WAEC & NECO
Over 3,900 Lesson Notes
Offline Support - Learn Anytime, Anywhere
Green Bridge Timetable
Literature Summaries & Potential Questions
Track Your Performance & Progress
In-depth Explanations for Comprehensive Learning